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```Unit 6 - Students Absolutely Must Learn…
Weekly Activity 11: RLC Circuit - AC Source





How to use the sinusoidal solutions of the RLC circuit components and what they mean
(including phase shifts).
How to relate the RLC circuit current to other circuit parameters especially the circuit
impedance.
How conservation of energy and calculus explain the /2 phase shifts.
How to determine the resonance features of the RLC circuit and how the various
parameters affect resonance.



How the RLC circuit's resonance may be used in technology.
How to modulate a carrier wave and envelope.
1
Staple the lab report, then graphs, and finally worksheets together. Please put
next lab meeting following the completion of the unit.
Unit Lab Report [50%, graded out of 25 points]
Write a separate section using the section titles below (be sure to label these sections in your
report). In order to save time, you may add diagrams and equations by hand to your final
printout. However, images, text or equations plagiarized from the internet are not allowed!
Remember to write your report alone as collaborating with a lab partner may make you both
guilty of plagiarism. Pay close attention to your teacher for any changes to these guidelines.

Title [0 points] – A catchy title worth zero points so make it fun.

Concepts & Equations [9 points] – {One small paragraph for each important concept, as
many paragraphs as it takes, 2+ pages.} Go over the lab activities and make a list of all
the different concepts and equations that were covered. Then simply one at a time
write a short paragraph explaining them. You must write using sentences & paragraphs;
bulleted lists are unacceptable.
Some example concepts for this unit report include (but are not limited to):
 Explain the meaning and use of the four time dependent equations describing
the component voltages.
 Explain the meaning and use of the resonant frequency.
 Explain how energy is stored in a capacitor and how energy is stored in an
inductor.
 Explain how to experimentally find an unknown inductance using an RLC circuit.
 Derive the equation for the resonant frequency.
 Derive the phase shifts of the inductor and capacitor using calculus.
2  t shift
 Derive why source 
.
T
 How wave modulation works.
 How the two-solenoid circuit was used to transmit a signal.
 How
 Any other equations that were used in the activities will need explained.
 Any other specific TA requests:

____________________________________________________________

____________________________________________________________

____________________________________________________________
2


____________________________________________________________
Selected In-Class Section [6 points]: {3-5 paragraphs, ~1 page}
This week's selection is: Weekly Activity 12, In-Lab Section 3
Write a "mini-report" for this section of the lab manual. Describe what you did
succinctly, and then what you found accurately. Then explain what the result means
and how it relates to some of the concepts in the previous section. You must write using
sentences & paragraphs; bulleted lists are unacceptable.
o Procedure: Do not provide a lot of specific details, but rather you should
summarize the procedure so that a student who took the course a few years ago
would understand what you did.
o Results: Do not bother to rewrite tables of data, but rather refer to the page
number on which it is found. State any measured values, slopes of ilnes-of-bestfit, etc. Do not interpret your results, save any interpretation for the discussion.
o Discussion: Analyze and interpret the results you observed/measured in terms of
some of the concepts and equations of this unit. It is all right to sound repetitive
with other parts of the report.

Open-Ended / Creative Design [6 points] – {3-5 paragraph, ~1 page} Choose one of the
open-ended experiments from the two weekly activities to write about. Describe your
experimental goal and the question you were trying to answer. Explain the ideas you
came up with and what you tried. If your attempts were successful, explain your results.
If your attempts resulted in failure, explain what went wrong and what you would do
differently in the future. You must write using sentences & paragraphs; bulleted lists are
unacceptable.

Graphs [4 points] - {attach to typed report} Graphs must be neatly hand-drawn during
lab and placed directly after your typed discussion (before your quizzes and selected
worksheets). Your graphs must fill the entire page (requires planning ahead) and must
include: a descriptive title, labeled axes, numeric tic marks on the axes, unit labels on the
axes, and if the graph is linear, the line of best fit written directly onto the graph.
3
Thoroughly Completed Activity Worksheets [30%, graded out of 15 points]

Week 11 In-Class [7 points]: Pages assigned to turn in:
_TA signature page, Post-lab pages, ____________________________________
___________________________________________________________________

Week 12 In-Class [8 points]: Pages assigned to turn in:
_TA signature page, _____________ ____________________________________
___________________________________________________________________
The above lab report and worksheets account for 80% of your unit grade. The
other 20% comes from your weekly quizzes, each worth 10%. These will be
entered into D2L separately.
4
Weekly Activity 11: RLC Circuit - AC Source
Pre-Lab
!
You must complete this pre-lab section before you attend your lab to prepare
for a short quiz. Be sure to complete all pages of the pre-lab.
Continue until you see the stop pre-lab picture:
Just as the RC circuit with a sinusoidal source had sinusoidal voltages containing phase shifts, so
too does the RLC circuit with sinusoidal source (that you will need to memorize rather than
derive):
V (t ) source  Vsource
sin drivet  shift   
amplitude
V (t ) resistor  Vresistor sin drivet 
amplitude


V (t ) capacitor  Vcapacitor sin  drivet  
2
amplitude




V (t ) inductor  L Vinductor sin  drivet  
Z amplitude 
2
These equations are graphed in the following picture:
5
This graph shows that the resistor voltage has no phase shift as its graph goes through the
origin. The capacitor voltage is phase shifted by /2 ahead of the resistor voltage (the capacitor
voltage leads the resistor voltage by 90o). The inductor voltage is phase shifted by /2 behind
of the resistor voltage (the inductor voltage lags the resistor voltage by 90o). Finally, the source
voltage is the negative sum of the other components VS (t )  VR (t )  VC (t )  VL (t ) due to
conservation of energy VS (t )  V R (t )  VC (t )  V L (t )  0 . The phase shift of the source is more
complicated depending on the parameters of the circuit: inductance L and capacitance C. This
formula is provided later.
In a circuit where an inductor, resistor and capacitor (RLC) are connected in series and driven by
a sinusoidal voltage source, the properties of the sinusoidally driven RC circuit and RL circuit (in
the solenoid lab) that you studied previously combine in a straightforward manner. Let’s
summarize the results for the RLC circuit that you should already suspect.
The voltage across each component oscillates at the same frequency as the driving frequency of
the source, drive. The properties of inductor and capacitor are frequency dependent, which
makes the circuit respond differently to different driving frequencies. This causes there to be a
specific frequency where the circuit has the largest possible current. This is the resonant
frequency,  resonant .
6
The equations for the circuit can be modified to be a little more accurate by substituting the
component amplitudes in terms of the source amplitude and component parameters. Some
students prefer to work with these, but an understanding of how to use both sets is critical.
V (t ) source  Vsource
sin drivet  shift   
amplitude
V (t) resistor 

R
Vsource sin  drivet 
Z amplitude
V (t) capacitor 
V (t )inductor 

The Resistor Voltage Equation:

 
Vsource sin  drivet  
Z amplitude 
2 
C
L


Vsource sin  drivet  
Z amplitude 
2
V (t) resistor 
R
Vsource sin  drivet 
Z amplitude
The resistor voltage is chosen as the voltage reference because it is the only ohmic device and
therefore can be used to determine the current in the circuit:

I(t) 
1
Vsource sin  drivet  ,
Z amplitude
where Z [ohm] is the total circuit impedance,
Z  R 2   L  C  .
2

This means that the current amplitude is simply the amplitude of the applied voltage divided by
the total circuit "resistance",
 the impedance:
I current
amplitude

Vsource
amplitude
Z
7
The circuit impedance acts like the total "resistance" of the circuit. The smaller Z is, the larger
the current amplitude. (Remember the current is oscillating back and forth in this 'AC' circuit.)
Z has two interesting parameters that each have the SI unit of [ohm], C [] is the capacitive
1
reactance  C 
and L [] is the inductive reactance L  driveL . These parameters
 driveC
depend on the driving frequency of the circuit  drive . This means that if you change the driving
frequency, you will change the total circuit impedance Z, and thus change the amount of
 this frequency dependence explicitly:
current that can flow in the circuit I. Substitution shows

1 

Z  R 2   drive L 

C
drive 

2
¿
0-1
Is the current amplitude in the circuit increased or decreased when the
difference between the capacitive reactance and inductive reactance is
increased?
When is Z (and thus current) a maximum?
Note that
2
2
 L  C   C   L  .

Driving Source Voltage Equation: V (t )source
 Vsource sin drivet  source   
amplitude
The source phase shift is given by
voltage will be in phase
L  C because tan10  0 .
  C 
source  tan1 L
.
(albeit

negative)
R

with
the
This shows that the source
resistor
voltage
when
¿

0-2

When the circuit is driven at the resonant frequency  resonant , this causes  L   C .
What is the phase shift of the voltage source when the circuit is driven at the
resonant frequency? This important fact is used in lab to find the resonant
frequency.
8
The Capacitor Voltage Equation:

 
Vsource sin  drivet  
Z amplitude 
2 
C
V (t) capacitor 
The capacitor voltage is phase shifted to lag the resistor voltage by 90 o. Note that the ratio of
the capacitor's "resistance" to the total circuit "resistance" determines the factor multiplied to
the source voltage to determine
the capacitor's voltage amplitude.

¿
0-3
Explain why it is possible for  C to be larger than Z so that Vcapacitor is larger than
amplitude
Vsource . Examine
Z  R 2   L  C  carefully and note that the effects of  C
2
amplitude
can be cancelled out by another component.

The Inductor Voltage Equation:
V (t )inductor 
L


Vsource sin  drivet  
Z amplitude 
2
The inductor voltage is phase shifted to lead the resistor voltage by 90o. Note that the ratio of
the inductor's "resistance" to the total circuit "resistance" determines the factor multiplied to
the source voltage to determine the inductor's voltage amplitude.
¿
0-4
What is the phase relationship between the inductor and capacitor voltages,
V L, amp and V C, amp ?
9
The above picture shows the sinusoidally driven RLC circuit at an instant in time when the
source voltage is positive in the upward direction. That causes the current to flow counter
clockwise. Each component of the circuit has a plus and minus sign to show on which side the
component voltage is higher or lower.
 The voltage drops as the current crosses the resistor.
 The capacitor has positive charge building up on the top plate as the current piles up
there and an equal amount of negative charge on the bottom plate.
 The inductor has the direction of its signs reversed from the other circuit components.
This is because the inductor's induced voltage is a back EMF (voltage) and could also be
written as  L (t ) . Remember that for an inductor (solenoid) the changing current
produces a changing magnetic field within the solenoid that in turn creates a back
voltage (EMF). A pneumonic device might be to think of an inductor as a grumpy person
who doesn't like change so it creates a voltage to try and stop the changing voltage of
the source.
¿
0-5
The above picture is a very useful way to think about and analyze circuits, but it
is not very honest. It seems to imply that all the components have +/- voltage
separation at the same time. Use your knowledge of the phase relationships
between the components to explain what really happens.
10
Sometimes these four boxed (important) voltage equations are rewritten using the fact that
Iamplitude 
Vsource amplitude
:
Z
V (t )source  Z  I amplitude sin drivet  source   

V(t)resistor  R Iamplitude sin drivet 


¿ 0-6

 
V (t)capacitor  C  Iamplitude sin  drivet  

2 

 
V (t) inductor   L  Iamplitude sin  drivet  

2 
If R = 10 [, L = 10 [H], C = 10 [F], Vsource amplitude = 10 [V] and fdrive = 10 [Hz], find
the current amplitude I amp and the voltage amplitudes of the inductor and

capacitor, V L, amp and V C, amp .
¿
0-7
Use the parameters and results of the previous question to write an equation
for the time dependence of the current I (t ) . Be sure to get the phase relation of
the current to the resistor voltage correct.
11
The above picture is constructed assuming that  C is larger than  L . This causes V C, amp to be
larger than V L, amp . Note that V R, amp is larger than the other two voltage amplitudes, which
indicates that R is larger than  C and  L . Thus because R >  C >  L you can see that V R, amp >
V C, amp > V L, amp . Always keep in mind that  C and  L depend on the driving frequency  drive ,
C 
1
 drive C
and L  driveL .
Since the capacitor voltage lags the resistor voltage by /2 while the inductor voltage leads the

resistor voltage
by /2, the capacitor and inductor voltages themselves are 180o out of phase
(/2 + /2 = ). This means that V C (t ) and V L (t ) always oppose each other and that their
combined voltages subtract from each other.
¿
0-8
Because Z  R 2   L  C  ,  C and or  L can be larger than Z (but R is always
smaller than Z). If R  3 [] ,  C  10 [] ,  L  6 [] , and I amp  10 [A] calculate Z,
and then all the voltage amplitudes, VS, amp , V R, amp , V C, amp , V L, amp , and then order

the
voltage amplitudes from largest to smallest. (Notice that the source voltage
amplitude is not the largest.)
2
12
In this example, the capacitor voltage amplitude is larger than the inductor voltage amplitude.
This causes the source voltage to reach it’s amplitude a little after the resistor does. Recall that
  C 
source  tan1 L
 and since C is larger than L, you obtain source < 0. This means that the
 R 
source voltage lags the resistor voltage. You can see this qualitatively by seeing that the source
voltage peaks between the resistor and capacitor peak voltages.

At driving frequencies where L > C, you find that VL > VC, and this causes the source voltage to
Finally, imagine that you are able to adjust the source frequency while leaving V source,amplitude
V
constant. Since Iamplitude  source amplitude , the current in the resistor can be maximized by
Z
minimizing Z. Remember that Z depends on drive since both C and L each depend on drive.
Since  Z  R 2   L  C  , Z is minimized when C is equal to L so that L – C, = 0. This
 = R.
gives Zminimum
2
 The driving frequency at which L – C, = 0 occurs is called the resonant frequency. This can be
1
found by setting C equal to L so that
  driveL. Some algebra (that should appear in
 driveC


 drive
resonance

1
LC
13
Imagine that you have an RLC circuit being driven sinusoidally at resonance. The picture below
shows that when the circuit is driven at the resonance frequency, C equal to L implies that
the capacitor and inductor voltages have equal magnitude. Since they are 180 o phase shifted,
their voltages are equal and opposite. So
VS ( t )  V R ( t )  V C ( t )  V L ( t )  0

 

cancel
and thus
VS (t )  V R (t )  0
so that the voltage of the source and resistor also exactly equal and opposite. This also means
that the source voltage phase shift source is zero since the source voltage peaks at exactly the
same time as the resistor voltage and directly in between the inductor and capacitor peak
voltages.
The sinusoidally driven RLC circuit at resonance.
14
¿
0-9
Compare the voltage amplitudes of the inductor and capacitor, V C, amp and V L, amp ,
when the circuit is driven at the resonant frequency  resonant ?
¿
0-10
Compare the circuit impedance Z to the resistor's resistance R when the circuit
is driven at resonance.
¿
0-11
Compare the voltage amplitudes of the inductor and capacitor, VS, amp and V R, amp ,
when the circuit is driven at the resonant frequency  resonant ?
¿
0-12
If the capacitor is removed from the circuit so that one has a sinusoidally driven
RL circuit, what driving frequency  drive minimizes the total circuit impedance Z
and thereby maximizes the circuit current? Use your answer to justify why an
inductor is sometimes called a "low pass filter".
15
Practice the math for an example of a sinusoidally driven series RLC circuit.
¿
0-13
Calculate each of the basic RLC circuit parameters in SI units though not
necessarily in the order given (SI units):


C
L
Z
VR,amplitude

VC,amplitude

VL,amplitude



source
Iamplitude

16
¿
0-14
Next write a time dependent equation for each quantity below using the
numerical results from the previous question.
Vsource(t)


VR (t) 




VC (t)


VL (t) 




I(t)

Qcap (t)




¿
0-15
What is the resonant driving frequency f resonant of this circuit?
17
18
In-Lab Section 1: measuring with two oscilloscope channels
In electronics, you may use the oscilloscope to simultaneously measure the voltage of two
adjacent components. Connecting your inductor, capacitor and resistor in series should be
done with some consideration of which components you will measure with your oscilloscope
since you only have two channels to measure with. The three circuit configuration examples
below and the questions that follow are used to illustrate this point
¿
1-1
In circuit A, which pairs of components can be separately measured on the
oscilloscope with a middle ground technique? (You are free to choose the
location of the oscilloscope ground.)
¿
1-2
In circuit B, which pairs of components cannot be separately measured on the
oscilloscope with a middle ground technique? (You are free to choose the
location of the oscilloscope ground.)
¿
1-3
In circuit C, which pairs of components can be separately measured on the
oscilloscope with a middle ground technique? (You are free to choose the
location of the oscilloscope ground.)
!
Depending on which two components you wish to measure, you often have to
rearrange the actual components.
19
¿
1-4
Sketch a circuit configuration along with the proper placement of the three
oscilloscope leads (red 1, red 2 and ground) in order to be able to separately
measure the time dependence of the solenoid and resistor simultaneously.
¿
1-5
Sketch a circuit configuration along with the proper placement of the three
oscilloscope leads (red 1, red 2 and ground) in order to be able to separately
measure the time dependence of the source and resistor simultaneously.
20
In-Lab Section 2: calculus explains /2 phase shifts
The explanation of the 90o phase shifts of the inductor and capacitor voltages from the resistor
voltage can be explained using calculus.
Conservation of energy for the sinusoidally driven RLC circuit gives:
Vsource (t )  VL (t )  VR (t )  VC (t )  0 .
Substituting Ohm’s Law VR (t)  I(t)  R , the definition of inductance VL (t)  L
definition of capacitance VC (t) 

Q(t)
gives:
C
dI(t)
and the
dt
dI (t )
Q (t )
Vsource (t )  L
 R  I (t )  
0

.
dt
C

 resistor term

inductor term
capacitor term
First notice that first two terms on the right hand side of the equation show that the time
dependence of the inductor is related to the resistor by a time derivative,
dI (t )
Q (t )
Vsource (t )  L
 R  I (t ) 
0
dt
C

.
The inductor voltageis
proportional to the time
derivativeof the resistor voltage.
All time dependent quantities of the circuit are oscillating sinusoidally including Q (t ) and I (t ) .
Since the resistor voltage is described by a sine function, V (t) resistor  Vresistor sin  drivet , the
current is as well, I (t )  I amp sin  drivet  . Since
V L (t )  L
amplitude
dI (t )
 L  Iamp cos drivet  ,
dt
so the inductor voltage must be described by a cosine function. But the cosine function leads
the sine function by /2, so the inductor voltage leads the resistor voltage by  /2,


L  I amp cos drivet   L  I amp sin   drivet   .
2

Thus calculus has explained the 90o phase shift of the inductor voltage relative to the resistor
voltage.
21
Next remember the definition of current I(t) 
dQ(t)
to find that
dt
dI (t )
dQ(t )
Q (t )
Vsource (t )  L
R

0
.
dt 
dt
C





inductor term
resistor term
capacitor term
Notice that the second two terms on the right hand side show that the time dependence of the
capacitor voltage is related to the antiderivative of the resistor voltage,
Vsource (t )  L
dI (t )

dt
dQ(t ) Q(t )
R

dt
C

0
The capacitor voltageis
proportional to the antiderivative
of the resistor voltage.
Since the resistor voltage is described by a sine function, V (t) resistor  Vresistor sin  drivet , the
amplitude
capacitor voltage must be described by a negative cosine function, the antiderivative of the sin
function. But the negative cosine function lags the sine function by /2,


 cosdrivet   sin drivet  
2

So the capacitor voltage lags the resistor voltage by  /2.
¿
2-1
Use the fact that
dI (t ) d 2Q (t )

and use calculus explain in a different way why
dt
dt 2
the inductor voltage leads the resistor voltage by /2. Begin with
d 2Q ( t )
dQ(t ) Q(t )
Vsource (t )  L
 R

 0 and Q (t )  Q cos t  .
2
amp
drive
dt
dt
C

22
In-Lab Section 3: the undriven LC circuit
There is a related circuit that should also be mentioned, the LC circuit (undriven but with an
initially charged capacitor).
Imagine that the capacitor was initially charged and placed into this circuit. At the instant
before charge began flowing from the capacitor, there would be an electric field inside the
capacitor and no fields or current inside the inductor.
The electric field in the capacitor contains energy,
1
E energy of  CV 2.
2
capacitor
E-field
As charge begins to flow from the capacitor, energy stored in the electric field inside the
capacitor decreases. However, there is no resistor in the circuit to remove energy by creating
heat so the total energy of
the circuit must be conserved. So this energy must go somewhere.
As current begins to flow through the inductor, the solenoid will have a magnetic field created
inside it. The energy stored in the inductor’s magnetic field is
1
E energy of  LI 2 .
2
inductor
B-field
The maximum current will be reached at the moment the capacitor is completely discharged.
This means that the moment when all the charge has left the capacitor, all the energy stored in
the electric field inside the
 capacitor will have been transformed into energy stored in the
magnetic field inside the inductor,
E energy of  E energy of .
capacitor
E-field

inductor
B-field
23
After some time, the charge will flow back onto the capacitor (in the reverse direction) and the
E energy of  E energy of .
capacitor
E-field
inductor
B-field
And so it will go, energy swinging back and forth between the electric field of the capacitor and
the magnetic field of the inductor, but at all times the total energy is conserved

E total  E energy of  E energy of .
energy
capacitor
E-field
inductor
B-field
¿
3-1
Use the same analysis as in section to write a total voltage equation relating
d 2Q(t)

Q(t) and
for the undriven LC circuit. (Simply plug in zero for the missing
dt 2
components.)

¿
3-2
If you know how, you should prove to yourself that the solution to the
differential equation you found in the previous question results in an oscillation
frequency of LC circuit 
1
LC
.
¿
3-3
Quickly sketch a voltage vs. time graph below of the simultaneous inductor and
capacitor voltages for this circuit.
24
In-Lab Section 4: extra lab techniques
An RLC circuit has a resonant driving frequency fresonance that maximizes the circuit current
amplitude. Therefore, a plot of the current amplitude vs. driving frequency will show peak
current amplitude at the resonant driving frequency. However, this graph can appear different
depending on whether a large or small resistance is used in the circuit:
The sharpness of the resonance peak is measure by the quality factor Q. The larger the quality
factor, the sharper the peak. Q is found to be given as
Q
 resonance  L
R
.
¿
4-1
What are the units of Q?

¿
4-2
Say you were building a circuit that needed to operate at many possible driving
frequencies, and you wanted to avoid shorting out the components with a large
current. Should you use a larger or smaller resistance?
¿
4-3
Say you were building a circuit that needed to operate at one specific frequency
(near resonance), and you wanted to maximize the current through the circuit
to do some work. Should you use a larger or smaller resistance?
25
In today’s lab you usually want to use as small of a resistance as possible (while still being safe)
in order to obtain a nice sharp peak in the current amplitude. This will make it easier to find the
resonance frequency. However, it is not always desirable to have a large quality factor. In
some mechanical systems (bridges, buildings, etc.) a pronounced response to a driving
frequency can cause destruction.
The most accurate way to find fresonance is to utilize the fact that at resonance, VR(t) and Vsource(t)
are exactly in phase with each other with equal amplitudes. You should place each of these
voltages on your oscilloscope channels and examine an XY formatted display. The resonance
frequency is easily found because you will see an ellipse when VR(t) and Vsource(t) are out of
phase and a diagonal line when they are in phase. You see a straight line when they are in
phase because both voltages must reach zero simultaneously.
Note that if R is chosen small enough, the resistance of the very long wire of the solenoid may
become an appreciable resistance of the circuit. When this occurs, the amplitude of the
resistor voltage will be smaller than the source voltage amplitude so that you will see a
different angle of tilt (not 45o) of the in phase resonance line shown above. This is because
Vsource  VR  Vresistanceof  VL  VC  0 .
real solenoid
26
In-Lab Section 5: specific oscilloscope measurements
Construct an RLC circuit driven by a sinusoidal driving voltage of Vsource
 5 [V] with
amplitude
The frequency and
f drive  10,000 [Hz] , R  1 [k ] , C  0.1 [ F] , and L  50 [mH] .
resistance values were not randomly selected, but were determined so that R,  C and  L are
of the same order of magnitude (check if you like). This means that V R, amp , V C, amp and V L, amp
will also be of the same order of magnitude. This "cheat" is only for this, your first time so you
learn the measurement techniques better. Later you will need to be able to analyze two
components on your oscilloscope when their voltages are drastically different.
¿
5-1
Why must you hook up the entire circuit before setting the source voltage on
the function generator? Why not just hook the function generator directly to
the oscilloscope and set its properties first?
!
You must keep this driving frequency the same as you are answering different
questions (until told otherwise) because the properties of the circuit change
with driving frequency.
¿
5-2
It is often the case that components in a circuit become disconnected at bad
solder joints even though you cannot see this with the unaided eye. What can
you do experimentally to make sure that your RLC circuit is still conducting
current through all its connections without disconnecting the circuit? {Hint:
what information can the ohmic component give you?}
¿
5-3
Quickly measure the voltage amplitudes of each component separately using a
amplitude is 5 [V] first.
27
¿
5-4
Now use two-channel observations to simultaneously measure the resistor
voltage and the inductor voltage to determine their respective phase shift.
¿
5-5
Now use two-channel observations to simultaneously measure the resistor
voltage and the capacitor voltage to determine their respective phase shift.
¿
5-6
Now use two-channel observations to simultaneously measure the inductor
voltage and the capacitor voltage to determine their respective phase shift.
Now use two-channel observations to simultaneously measure the resistor voltage and the
source voltage to determine their respective phase shift. Note that a phase shift of /2
corresponds to a shift along the time axis of ¼ of oscillation cycle T. Therefore, if t shift is the
amount of time lag or time lead that you measure between the resistor amplitude and the
source amplitude, you can set up the following equivalent ratios to find the phase shift of the
source source,
t shift

 source .
1   
 T   
4  2 
This gives
source 

2  t shift
  drive  tshift .
T
Another way to think about how to measure the source voltage phase shift is to examine the
extra amount of time it would take for the resistor voltage to come into phase with the source
voltage if the source voltage was frozen in time:

sin  drivet  t shift   sin  drivet  source.
For these sine functions to be equal, their arguments must be equal so that you find
source  drive  tshift .
¿ 5-7

Use this equation to experimentally determine the source voltage phase shift
source from the resistor voltage.

28
¿
5-8
Theoretical calculations: Use the labeled values for inductance, capacitance and
resistance that you have in your circuit as well as your actual driving frequency
to calculate the following RLC circuit parameters in SI units.

C
L
Z


resonance
fresonance
¿
5-9
Comparing theory and observation: Use your results from the previous
question along with your observed source voltage amplitude (should be close to
5 [V]) to theoretically calculate the voltage amplitudes of the resistor, inductor
and capacitor in SI units. Then compare each of these theoretically computed
29
Presistor  I Vresistor 
Power: Since
Vresistor 2
R
,
the power converted to heat by the resistor oscillates in time as a squared sine function:
PR (t ) 
V R2 (t )
R

V R,2amp
R
sin 2 drivet  .
The average value of a squared sine function over a complete oscillation cycle is 0.5:


average sin(  drivet  
2
1
.
2
This means that the power consumed by the resistor averaged over time is

PR, average 
V R,2amplitude
2R
.
¿
5-10
Use this formula to find the average power loss of your circuit in SI units.
¿
5-11
When the driving frequency is at resonance, the voltages of the source and
resistor are equal and in phase. Use this fact to search the range of driving
frequencies for the resonant driving frequency using the technique described in
section 4 of this lab. Compare this observed value to your predicted value in 57.
30
¿
5-12
Collect data of Iamplitude vs. fdrive and record below. Then graph the data on
separate graph paper. A poor example with a larger resistance is shown to help
guide you.

31
32
In-Lab Section 6: authentic assessment
One solenoid and each of the four capacitors on the capacitor board allows you to construct
four unique RLC circuits (not counting the resistance possibilities). You need to find the four
resonant frequencies of each of these RLC circuits using the method described in section 4 on
the phase relationship between the resistor and source voltages. Use a the same solenoid as a
different group so that you can compare answers.
!
The inductor and capacitor voltage amplitudes can be larger than the source
voltage amplitude. Voltages over 20 [V] can cause nasty shocks, and very large
currents can be generated near resonance in circuits with a large quality factor
Q. You should approach any circuit with care, not touching the components
while the circuit is powered. In designing a circuit, one must either use a large
resistance to limit the dangerous currents near resonance, or have a firm
theoretical understanding of the maximum currents that the circuit can
produce, while still proceeding with caution.
¿ 6-1
Show a student in a different group that you can successfully measure the
resonance frequency of an RLC circuit (four circuits, compare your data with
theirs). Once you are successful, have them sign below. Note: if someone is
"Yes, I have seen this student successfully the resonance frequencies of four RLC
circuits. Their results match mine. Either they are doing it right or we are both
wrong in the same way!"
Student Signature:___________________________________________________
33
In-Lab Section 7: open-ended / creative design
Pretend that you are an old-timey inventor. You have already discovered the coiled wire device
(now called inductor) and the metallic parallel plates device (now called capacitor). Resistors
were discovered by a competitor of yours. It is still a sore subject around the workshop.
You have begun to combine these new components together in order to observe their
compound behavior. You already found that when the inductor and capacitor are combined in
series and driven sinusoidally, the circuit produces a special resonant frequency where the
current in the circuit is a maximum. This means that civilization can now make electronics that
“select” specific driving frequencies and “suppress” frequencies far from resonance.
Now you try combining the inductor and capacitor in parallel and drive them sinusoidally (see
above figure). What electrical possibilities might this new compound device hold for the future
of mankind? Well, use your anachronistic oscilloscope to find out. Try to deduce why this
circuit may be said to have an “antiresonant frequency” and see if you can find it.
You are allowed to "cheat" by talking to other groups for ideas, but are not allowed to "cheat"
Below you are given three prompts:
hypothesizing/planning, observations/data,
calculations/conclusion. Your job is to figure out the answer using these prompts as your
problem-solving model. In the event that you should run out of time, you may not discover the
correct answer, but you should make an attempt at each prompt. Grades are based on honest
effort.
Your open-ended solution should probably include some of the following items: sketches of
circuit diagrams, tables of data, calculations, recorded observations, random ideas, etc.
Write at the prompts on the next page.
34
¿
7-1
hypothesizing/planning:
¿
7-2
observations/data:
¿
7-3
calculations/conclusion
I, the physics 241 laboratory TA, have examined this student's Weekly Activity pages and found
them to be thoroughly completed.
!
TA signature: _______________________________________________________________
35
Post-Lab: RLC Circuit - AC Source - REVIEW PRACTICE CAPACITORS
!
You must complete this post-lab section after you attend your lab. You may
work on this post-lab during lab if you have time and have finished all the other
lab sections.
¿
X-1
A 1.5 [V] battery is connected to two conducting shapes that are not touching
(i.e. a capacitor), determine the voltage difference between the following point
pairs:
i) Va - Vc =
ii) Va - Vb =
iii) Vd - Vb =
¿
X-2
Explain which picture above correctly describes the true charge configuration
for this capacitor. {Hint: What must the electric field be inside a conductor?}
36
¿
X-3
The above circuit has switch 1 (S1) and switch 2 (S2) initially open. At t=0.0 [s],
switch 1 is closed allowing current to flow around the outer part of the circuit
and begin charging the capacitor. At t=1.0 [s], switch 1 is opened again. At
t=2.0 [s], switch 2 is closed (while switch 1 stays open). At t=2.2 [s], switch 2 is
reopened.
a) At t=0.5 [s], find VC, QC, and IR2.
b) At t=1.5 [s], find VC, QC, and IR2.
c) At t=2.1 [s], find VC, QC, and IR2.
d) At t=100.0 [s], find VC, QC, and IR2.
37
38
Pre-Lab
!
You must complete this pre-lab section before you attend your lab to prepare
for a short quiz. Be sure to complete all pages of the pre-lab.
Continue until you see the stop pre-lab picture:
Subsection 0-A
Reminder of how to measure an unknown capacitance C with an RC circuit:
The component voltage amplitudes of an RC circuit driven sinusoidally with angular frequency
drive are given by
VC,amp 
C
Z
VS, amp
and
VR, amp 
R
VS, amp .
Z
These component voltage amplitudes are equal when
C  R .  C
changes with different
driving frequencies, so if you can find the right driving frequency, you can get
VC, amp  VR, amp . Substituting R  C 
gives
C
C  R
so that
1
 driveC
and solving for C
1
R 2  fdrive .
Remember this is only true when the voltage amplitude across the resistor is equal to the
voltage amplitude across the capacitor. Therefore, to find an unknown capacitance C, measure
R and find the driving
frequency where VC, amp
 VR, amp .
39
Reminder of how to measure an unknown inductance L with an RL circuit:
In an RL circuit driven sinusoidally with angular frequency drive,
VL,amplitude 
and
VR,amplitude 

These
component
voltage
amplitudes
R   L  driveL and solving for L gives

L

L
Z
Vsource,amplitude
R
Vsource,amplitude .
Z
are
equal
when
L  R.
Substituting
R
2  fdrive .
Remember this is only true when the voltage amplitude across the resistor is equal to the
voltage amplitude across the inductor. Therefore, to find an unknown inductance L, measure R
and find the driving frequency where V L, amp

 VR, amp .
Reminder of how to measure the resonant frequency resonance of an RLC circuit:
In a sinusoidally driven RLC circuit, there is a driving frequency at which the current in the
resistor is maximized (i.e. absorbing the most power from the oscillating source). This happens
when the total circuit impedance
 L   C 
2
2
 0.


Z  R 2   L  C  is minimized. This occurs when
The driving frequency at which
L  C  0
can be found by setting C equal to L so that
 drive
resonance

1 
.
LC


occurs is called the resonant frequency. This
1
 driveC
  driveL.
Solving gives
40
Reminder the resistor's affect on the RLC circuit:
A plot of current vs. driving frequency in an RLC circuit has a maximum at the resonant
frequency. This is useful because an RLC circuit can be tuned to a specific resonant frequency
by adjusting its C or L. If you lower the resistance of an RLC circuit and retake your
measurements, the resonant frequency won’t change (since L and C don’t change), but you will
measure a higher quality factor:
The smaller the resistance, the sharper this peak gets. This is useful because an RLC circuit with
low resistance only “reacts to” frequencies very near resonance while ignoring other driving
frequencies. A radio can be made using a low-resistance RLC circuit that responds only to a
specific radio frequency electromagnetic wave. How could you make the peak as sharp as
possible in an RLC circuit for use in a radio? Take the resistor out so that only the tiny
resistance of metal wires is present.
Reminder of how to observe fresonance using an oscilloscope:
The most accurate way to find fresonance is to utilize the fact that at resonance, VR(t) and Vsource(t)
are exactly in phase with each other with equal amplitudes. You should place each of these
voltages on your oscilloscope channels and examine an XY formatted display. The resonance
frequency is easily found because you will see an ellipse when V R(t) and Vsource(t) are out of
phase and a diagonal line when they are in phase. You see a straight line when they are in
phase because both voltages must reach zero simultaneously.
41
Subsection 0-B
Modulating High Frequency Waves with Low Frequency Waves:
Imagine that you want to transmit a 1,500 [Hz] sound wave from one place to another. Clearly
you could play the sound and have its vibrations be carried in the air as
compression/rarefaction waves. But for long distance transport, you may want to turn the
'sound' into an electronic signal which can be transported with less energy loss along a wire.
But if we need to transport the sound to many people (radio) or someone at a changing
location (walkie-talkie), we can take advantage of our knowledge of electromagnetism and turn
the sound wave into an electromagnetic wave and send the electromagnetic wave through the
air. Of course an electromagnetic wave traveling through the air moves at the speed of light
because and electromagnetic wave is light. We can turn our sound wave into light to transport
it.
First investigate transporting sound as oscillating magnetic fields in solenoids, from one
solenoid (the transmitter) to another solenoid (the receiver). The two solenoids are not
connected in any way so that the oscillating magnetic field inside one solenoid must be made to
d B
oscillate within the other solenoid to utilize Faraday’s Law of inductance,  L  
, a
dt
process called mutual inductance.
Unfortunately, this wave is alternating much too slowly to induce a large voltage in the
receiving
solenoid.
Remember
the
equation
for
mutual
inductance,
dI circuit 1
M1 to 2 , where M is a constant that describes how much the solenoids
dt
overlap. If the current doesn’t oscillate rapidly enough, then V induced is very small.
V
induced
in circuit 2

in circuit 2
42


You might say, "gee, I wish this wave oscillated more quickly to cause a bigger induced voltage
in the receiving solenoid." But then it wouldn’t be the same sound pitch that you wanted to
hear in the first place! Still, it sounds like something you might say since you understand that to
utilize Faraday's law, you need rapidly changing magnetic fields.
Next examine a wave that oscillates quickly at radio wave frequency or slightly sub-radio
frequency.
This isn’t the frequency you want to hear (you are not able to!), but it does oscillate so quickly
as to create a large induced voltage in the receiving solenoid. I.e., it oscillates quickly enough to
be transmitted into the receiving circuit through the mutual inductance of the “transformer”
(overlapping solenoids).
So what to do? There is a low frequency wave that carries the information we wish to transmit
but can't, and there is a high frequency wave that we don't care about that is easily
transmitted.
The solution is to combine the two waves together by multiplying them. This modulated wave
has the properties of both waves: it carries information about the audio frequency component
and it oscillates quickly enough to generate a highly induced voltage in the receiver circuit. The
following picture shows what a modulated wave looks like.
43
The envelope wave is the low frequency oscillation while the high frequency oscillation is called
the carrier wave. Together they are the modulated wave.
¿
0-B-1
Why are we interested in using a low frequency envelope wave? What kind of
useful information might it hold?
¿
0-B-2
Why do we need to use a rapidly oscillating carrier wave? What purpose could
it serve?
44
In-Lab Section 1: introducing the equipment
In today’s lab, we would like to transmit a modulated sound wave transmitted by one solenoid
into another “receiving” solenoid. We will use a capacitor in the second “receiving” circuit to
make an RLC “receiving” circuit. By changing the capacitor of the “receiving” circuit, we can
adjust its resonant frequency. Therefore, we will be able to “tune” our “receiver” to a
Now you would like to listen to your transmitted wave. But there is a huge problem.
Whenever the wave is positive, it causes an upward force on the speaker, and whenever it is
negative it causes a downward force on the speaker. The modulated wave is oscillating up and
down with the rapid radio frequency, much too fast for the speaker to respond to. It just sits
there quivering.
The trick is to add a diode to the output. Remember that a diode is a quantum mechanical
component that only allows current to flow in one direction once a turn-on voltage has been
reached (determined by the semiconductor band gap energy). This will allow only positive
voltage to reach the speaker.
The speaker now gets pushed out a maximum distance at the maximum amplitude of the pulse
and relaxes at the minimum.
¿
1-1
Would you be able to hear the speaker if the direction of the diode in the circuit
45
!
Never insert the speaker directly into your ear without first verifying that the
sound level is safe.
Now you will experimentally use a sound wave to modulate a radio frequency wave.
First examine the equipment with two solenoids:
This device allows you to see how two solenoids interact via Faraday's law. There is a small
visible gap that separates the two different coils of wire. If you look carefully, you can see that
GND connects to the left side of the primary solenoid and ANT connects to the right side of the
primary. You can also see that A connects to the left of the secondary and B to the right. In
addition to these connections, there are some junctions sticking out of the top of the solenoids.
These allow you to use smaller portions of each solenoid if you want less inductance L.
We will always want to use the connections at ANT and A. This forces the current to travel in
the parts of the primary and secondary that are adjacent to one another.
46
For example, if you wanted to half the inductance of the secondary circuit shown above, you
would simply reconnect the B terminal to the midpoint of the secondary solenoid so that only
half of that solenoid is used (shown below).
47
¿
1-2
How many possible primary solenoid inductances are possible? How many
secondary? {Don't forget that the solenoids must remain immediately adjacent
to each other.}
Now examine the equipment with the continuously variable capacitance:
The variable capacitor is made of two sets of overlapping plates. One set can be rotated so that
it overlaps more or less with the other set of plates. When they maximally overlap the
capacitor has the highest capacitance it can provide. When the plates don't overlap at all, the
capacitance is zero.
In addition to the capacitor plates, there is a diode and speaker connector in parallel to the
capacitor. Remember that the diode eliminates half the oscillating voltage sent to the speaker.
48
Now let's be sure you can make the 'sound'. Clean your ear-speaker with alcohol and connect it
to the earplug on the capacitor assembly. Set your function generator to 3,600 [Hz]. Connect
your function generator to the A and B connections on the variable capacitor assembly as
shown below.
Slowly increase the output voltage of the function generator until you can hear the signal.
!
Ask for help if you have any doubts about the proper operation of these devices.
¿
1-3
Give yourself a hearing test. Find the upper end of the frequency range of your
hearing (or another student if you do not want to test your own hearing). It
may be significantly less than your classmates' if you are older or have suffered
hearing loss due to loud noises. Also note any intermediate ranges of hearing
loss. Do these frequencies correspond to the kinds of music you listen to (too
loudly)?
49
Speakers create sound waves in the air by mechanically oscillating a solid object that strikes air
molecules. For a given constant output voltage, cheaper speakers will not respond to all
frequencies with the same intensity. Really cheap speakers may even have a few resonant
frequencies where they mechanically 'rattle'.
¿
1-4
Does your simple speaker have any resonant frequencies (probably)? Find and
record the frequencies that make the speaker the loudest? There may be more
than one resonant frequency for your speaker since it is a complex mechanical
system (only very simple systems have one resonant frequency).
50
In-Lab Section 2: the modulated wave
The radio frequency (RF) modulator takes an input envelope wave and modulates it with a
rapidly oscillating carrier wave. If you do not input any envelope wave, only the pure carrier
wave will be output. Hook your RF modulator directly to the oscilloscope without the input
envelope wave from the function generator. Use a carrier RF wave of 500,000 [Hz].
¿
2-1
Use the oscilloscope to determine the period of the 500 [kHz] RF carrier wave.
Does this make sense since T 
1
?
f
Now input an audio-frequency signal of 1,000 Hz from your function generator into your RF
modulator (shown below). Use a carrier radio-frequency wave of 500,000 [Hz] to create a
modulated output wave of 1,000 Hz envelope waves surrounding a 500 kHz high frequency
wave.
¿
2-2
Calculate the period of the envelope wave using T 
1
?
f
Examine the output of this signal on your oscilloscope using two different time ranges so that
you can see the audio-frequency signal and then separately the radio-frequency signal. To do
this, choose a seconds-per-division setting for the time axis that is appropriate for the time
scale of the wave you wish to examine (i.e., use the wave’s period to select the seconds-perdivision). You may need to press the “run-stop” button to view the wave packets if they appear
smeared out on the oscilloscope.
51
¿
2-3
Sketch the appearance of the modulated wave at both the audio frequency
scale and then the radio frequency scale, and record the time scales used to
observe the waves. Notice that the RF generator does not necessarily produce a
clean RF output as was discussed earlier in this handout. Also, be sure you can
hear the envelope waves of the modulated signal in your speaker.
Audio Frequency
52
In-Lab Section 3: mutual inductance "radio"
Now you will use a sound wave to modulate a radio frequency wave, then use mutual
inductance to transmit the radio frequency wave to a separate RLC circuit, and finally listen to
the "broadcast" sound wave. Set the capacitor to its maximum setting.
¿
3-1
First determine the inductances LA and LB of your solenoids using the methods
described earlier. In other words, for each solenoid create an RL circuit driven
sinusoidally and monitor the voltage across the resistor and inductor as you
vary the driving frequency. Use the appropriate equation to solve for the
inductance L. Even with the 10  resistor, you will need to use large driving
frequencies as these inductances are very small.
¿
3-2
Determine the capacitance of the capacitor.
Choose your favorite audio frequency wave to broadcast and a carrier wave initially 550 [kHz].
Instead of trying to hear the broadcast wave in the secondary circuit, send the output to your
oscilloscope so you can see the broadcast wave.
53
¿
3-3
The secondary circuit is supposed to be an RLC circuit with a resonance
frequency. Where is the resistance and why would you want the resistance to
be as low as possible?
¿
3-4
Do you expect the secondary circuit to have a resonant frequency nearer to the
audio frequency or the carrier frequency?
¿
3-5
Calculate the resonant frequency fresonance of the secondary circuit.
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3-6
Adjust the carrier frequency until a maximum output voltage is determined on
the oscilloscope. If it does not match your previous answer, get help. Explain
why the carrier wave needs to be at the resonance frequency of the secondary
circuit to maximize transmission of the modulated wave.
Now replace the oscilloscope with an actual speaker and use your ear to determine that
maximum broadcast transmission occurs when the carrier wave is at the resonance of the
secondary circuit.
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3-7
Write directly onto the figure below to completely explain how this compound
circuit(s) works. You can use labels, text and arrows, and you do not need to
write in complete sentences.
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In-Lab Section 4: authentic assessment - real radio wave reception
Now replace the function generator and RF modulator with a real antenna as shown:
Change the secondary solenoid inductance (4 options) and the capacitance (continuous
adjustment) to find radio stations which cause the secondary circuit to have a maximum
current (large voltage on the oscilloscope). Use the oscilloscope to determine the broadcast
frequency. Once you find a radio station, swap the oscilloscope with the speaker and see if you
can hear the transmitted envelope wave.
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4-1
in a different group and see if you found the same stations.
"Yes, I have seen this student use an RLC circuit to find invisible sound-carrying
light waves. We then rocked out to some jamming tunes!"
Student Signature:___________________________________________________
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In-Lab Section 5: open-ended / creative design
Imagine that you are about to finish a very long semester of electricity and magnetism labs, and
that all you have left to do is a single open-ended creative design question.
Using two of the large ~1 [H] solenoids stored in the corner of the room, create your own radio
station. Specifically, find a way to transmit unmodulated sound waves from one solenoid
through the air and listen to the sound using a second solenoid not connected in any way to the
transmitting solenoid. Decide on what pitch to broadcast using theoretical predictions based
on the numerical values of the components you have on hand. Points are awarded for loudness
(not really).
At the following prompts, design an experiment to loudly broadcast unmodulated sound waves
using two large solenoids among other electronics components. Then implement your
You are allowed to "cheat" by talking to other groups for ideas, but are not allowed to "cheat"
Below you are given three prompts:
hypothesizing/planning, observations/data,
calculations/conclusion. Your job is to figure out the answer using these prompts as your
problem-solving model. In the event that you should run out of time, you may not discover the
correct answer, but you should make an attempt at each prompt. Grades are based on honest
effort.
Your open-ended solution should probably include some of the following items: sketches of
circuit diagrams, tables of data, calculations, recorded observations, random ideas, etc.
Write at the prompts on the following page.
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5-1
hypothesizing/planning:
¿
5-2
observations/data:
¿
5-3
calculations/conclusion
I, the physics 241 laboratory TA, have examined this student's Weekly Activity pages and found
them to be thoroughly completed.
!
TA signature: _______________________________________________________________
59
!
If you have finished the lab and have time remaining, you should review lab
techniques for the lab practical:
 Using the DMM to measure:
o voltage
o current
o resistance
 Building circuits from schematics:
o series
o parallel
 Operation of a DC power supply.
 Operation of a function generator:
o square and sin waves
o output frequency
o output voltage amplitude
o DC offset
 Oscilloscope general usage:
o middle ground vs bottom ground setups
o time scale
o voltage scale
o where V=0 is on screen
o triggering
o screen capture
o how to get it off French if done so by accident
 Oscilloscope measurements (V vs t):
o frequency
o period
o voltage amplitude
o current amplitude (via resistor voltage amplitude)
 Oscilloscope measurements (Vx vs Vy):
o RLC resonance
 Tricky oscilloscope measurements:
o capacitance
o inductance
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